A gas undergoes a polytropic process (PV1.2 = constant) from an initial state of 3.4 bar and 0.283 m3 to a final pressure of 8.2 bar. If 0.183 kg of gas is involved and 2.2 kJ of heat is transferred f

Step 1: Identify given values and convert units. Given: Initial pressure, p_1 = 3.4 bar = 3.4 × 100 kPa = 340 kPa Initial volume, V_1 = 0.283 m^3 Final pressure, p_2 = 8.2 bar = 8.2 × 100 kPa = 820 kPa Polytropic index, n = 1.2 (from PV^1.2 = constant) Mass of gas, m = 0.183 kg Heat transfer from* the gas, Q = -2.2 kJ (negative sign indicates heat leaving the system) Changes in kinetic and potential energy are negligible. Step 2: Calculate the final volume, V_2. For a polytropic process, p_1 V_1^n = p_2 V_2^n. V_2 = V_1 ((p_1)/(p_2))^1/n Substitute the values: V_2 = 0.283 m^3 (3.4 bar8.2 bar)^1/1.2 V_2 = 0.283 m^3 ((3.4)/(8.2))^0.8333 V_2 = 0.283 m^3 (0.41463)^0.8333 V_2 = 0.283 m^3 × 0.50611 V_2 = 0.14332 m^3 Step 3: Calculate the work done, W, during the polytropic compression. For a polytropic process with n ≠ 1, the work done is given by: W = (p_2 V_2 - p_1 V_1)/(1-n) Substitute the values: W = (820 kPa)(0.14332 m^3) - (340 kPa)(0.283 m^3)1 - 1.2 W = 117.5224 kJ - 96.22 kJ-0.2 W = 21.3024 kJ-0.2 W = -106.512 kJ The negative sign indicates that work is done on the gas (compression). Step 4: Apply the First Law of Thermodynamics to find the total change in internal energy, U. The First Law of Thermodynamics for a closed system with negligible changes in kinetic and potential energy is: U = Q - W Substitute the values for Q and W: U = (-2.2 kJ) - (-106.512 kJ) U = -2.2 kJ + 106.512 kJ U = 104.312 kJ Step 5: Calculate the specific change in internal energy, u. Specific internal energy is the total internal energy divided by the mass: u = ( U)/(m) Substitute the values: u = 104.312 kJ0.183 kg u = 569.99 kJ/kg The heat transfer is given in the problem statement. The change in specific internal energy of the gas is 570.00 kJ/kg. The heat transfer is -2.2 kJ. That's 2 down. 3 left today — send the next one.